Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{n - 5}{n^2 + 7n} \div \dfrac{10n - 50}{n^2 + 11n + 28} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{n - 5}{n^2 + 7n} \times \dfrac{n^2 + 11n + 28}{10n - 50} $ First factor the quadratic. $p = \dfrac{n - 5}{n^2 + 7n} \times \dfrac{(n + 7)(n + 4)}{10n - 50} $ Then factor out any other terms. $p = \dfrac{n - 5}{n(n + 7)} \times \dfrac{(n + 7)(n + 4)}{10(n - 5)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (n - 5) \times (n + 7)(n + 4) } { n(n + 7) \times 10(n - 5) } $ $p = \dfrac{ (n - 5)(n + 7)(n + 4)}{ 10n(n + 7)(n - 5)} $ Notice that $(n - 5)$ and $(n + 7)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ \cancel{(n - 5)}(n + 7)(n + 4)}{ 10n\cancel{(n + 7)}(n - 5)} $ We are dividing by $n + 7$ , so $n + 7 \neq 0$ Therefore, $n \neq -7$ $p = \dfrac{ \cancel{(n - 5)}\cancel{(n + 7)}(n + 4)}{ 10n\cancel{(n + 7)}\cancel{(n - 5)}} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $p = \dfrac{n + 4}{10n} ; \space n \neq -7 ; \space n \neq 5 $